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(C) Hybridization is a concept used to explain the geometry of molecules by mixing atomic orbitals. $(i)$ $BCl_3$: Boron $(Z=5)$ has ground state conf

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$(ii), (iii)$

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(C) Hybridization is a concept used to explain the geometry of molecules by mixing atomic orbitals. $(i)$ $BCl_3$: Boron $(Z=5)$ has ground state configuration $1s^2 2s^2 2p^1$. To form $3$ bonds,it promotes an electron to $2p$ to form $sp^2$ hybrid orbitals. $(ii)$ $NH_3$: Nitrogen $(Z=7)$ has $1s^2 2s^2 2p^3$. It uses $sp^3$ hybridization in the ground state to form $3$ bonds and hold a lone pair. $(iii)$ $PCl_3$: Phosphorus $(Z=15)$ has $3s^2 3p^3$. It uses $sp^3$ hybridization in the ground state to form $3$ bonds and hold a lone pair. $(iv)$ $BeF_2$: Beryllium $(Z=4)$ has $1s^2 2s^2$. It promotes an electron to $2p$ to form $sp$ hybrid orbitals. Therefore,$(ii)$ and $(iii)$ undergo hybridization in their ground state electronic configurations without needing promotion of electrons to higher energy subshells.

	Molecule	Hybridization	Geometry	Number of lone pairs of electrons on central atom
$I$	$SiH_4$	$sp^3$	tetrahedral	$0$
$II$	$BeCl_2$	$sp^2$	linear	$1$
$III$	$SF_4$	$dsp^3$	square planar	$1$
$IV$	$SnCl_2$	$sp$	linear	$0$

Choose the molecules in which hybridization occurs in the ground state?
$(i)$ $BCl_3$ $(ii)$ $NH_3$
$(iii)$ $PCl_3$ $(iv)$ $BeF_2$
The correct answer is

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Which of the following sets is correct based on the table provided below?
Molecule Hybridization Geometry Number of lone pairs of electrons on central atom
$I$ $SiH_4$ $sp^3$ tetrahedral $0$
$II$ $BeCl_2$ $sp^2$ linear $1$
$III$ $SF_4$ $dsp^3$ square planar $1$
$IV$ $SnCl_2$ $sp$ linear $0$

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Choose the molecules in which hybridization occurs in the ground state?$(i)$ $BCl_3$ $(ii)$ $NH_3$$(iii)$ $PCl_3$ $(iv)$ $BeF_2$The correct answer is